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Infinite series - Inverse trigonometry

  1. Aug 18, 2013 #1
    1. The problem statement, all variables and given/known data
    The sum of the infinite terms of the series
    [tex]\text{arccot}\left(1^2+\frac{3}{4}\right)+\text{arccot}\left(2^2+\frac{3}{4}\right)+\text{arccot}\left(3^2+\frac{3}{4}\right)+............[/tex]
    is equal to
    A)arctan(1)
    B)arctan(2)
    C)arctan(3)
    D)arctan(4)

    Ans: B

    2. Relevant equations



    3. The attempt at a solution
    The given series can be written as
    $$\sum_{r=1}^{\infty} \text{arccot}\left(r^2+\frac{3}{4}\right)$$
    Simplifying the term inside the summation, I rewrite it as
    $$\arctan\left(\frac{4}{4r^2+3}\right)$$
    I don't see how to proceed further. I am thinking of converting the above in the form ##\arctan\left(\frac{x-y}{1+xy}\right)##, rewrite it in the form ##\arctan(x)-\arctan(y)## and perform the summation but I can't see a way to do this.

    Any help is appreciated. Thanks!
     
    Last edited: Aug 18, 2013
  2. jcsd
  3. Aug 18, 2013 #2

    vela

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    I haven't worked it out, but I think you're on the right track. Try rewriting it as
    $$r^2 + \frac{3}{4} = \left(r^2 - \frac{1}{4}\right) + 1 = \left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right) + 1$$
     
  4. Aug 18, 2013 #3
    Wow. That's great. Thanks a lot vela! :smile:

    I did try rewriting 3/4 as 1-1/4 but then I didn't notice that it could simplified to (r-1/2)(r+1/2). :redface:
     
  5. Aug 19, 2013 #4

    verty

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    This is one of the most sadistic questions I've seen. Where are you getting these questions?
     
  6. Aug 19, 2013 #5
    From practice sheets. Do you want me to send the links to those practice sheets by PM? I don't know if its ok to post them here.
     
  7. Aug 19, 2013 #6

    verty

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    A PM will be fine, thanks. I haven't seen these type questions before, is all.
     
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